Re: Promotion of bare decorations & comparisons
Nate, P1788
<First, please read the announcement at the end.>
On 12 Jan 2013, at 15:10, Nathan T. Hayes wrote:
> John Pryce wrote:
>> On 3 Jan 2013, at 22:08, Nathan T. Hayes wrote:
>>>> You can construct an equally silly example with that alternative:
>>>> [1,2] \subseteq floor([0,6])
>>>> ...
>>>> = [1,2] \subseteq Empty
>>>> = false
>>>>
>>>> Now we have a false negative.
>>>
>>> I don't agree it is a false negative.
>>>
>>> Compressing the decorated interval ([0,6],def) into the bare decoration def
>>> means the user has explicitly indicated anything less than dac is an error.
>>>
>>> So returning false in this example is exactly what the user expects, since
>>> [1,2] cannot be a subset of any defined and continuous interval range of
>>> floor([0,6]).
>>
>> I think "Hmm" on this one. What does "any defined and continuous (dac) interval range
>> of floor([0,6])" mean? The only meaning I can see is "since floor() isn't dac on the
>> input [0,6], such a range doesn't exist; and if we insist on treating this nonexistent
>> thing as a set, it must be the empty set".
> That is exactly my view... so why the "Hmmm"?
Precisely because I think the argument I stated above, with which you are agreeing, makes no sense.
Above, a certain kind of set does not exist: that is, the class of sets having a certain property P is empty. You are arguing from that that you can take the empty set as being a set that has property P. That way madness lies.
>> ...I don't think users would expect this: that (A \subseteq f(B)) is true for
>> some B, but becomes false when they make B larger.
>
> ???
Recently Michel wisely reminded us of the sentence, now in §6.1, originally written by Vladik I think:
(*)
> The decoration system is designed in a way that naive users of interval arithmetic do not notice anything about decorations, unless they inquire explicitly about their values.
I do not know if the standard can follow that 100%, but I want to follow it as closely as possible. In a set-based system, if f is a function given by an expression and B is a set, and f(B) means any sensible way of evaluating f on B (exact range; interval evaluation in infinite or finite precision,...), then
B \subseteq B' implies f(B) \subseteq f(B').
Isotonicity. But your "???" indicates you want it to be false because of a certain interpretation of decorations. Will the "naive user" think that is compatible with (*)?
[The Kaucher group may do something different, since its intervals are not *primarily* sets, and its \subseteq is not *primarily* set-containment.]
This issue arose in the context of compressed interval arithmetic but IMO really is due to our not yet having decided how boolean operations (comparisons) work for "ordinary" decorated intervals. We need to sort that soon, but it's not a killer problem, and shouldn't scare anyone from voting for motion 42.
But to simplify the issues, I follow Jürgen's proposal (today) and REMOVE the compressed arithmetic subclause from consideration under the main part of motion 42. I will discuss with the chair how this is best done.
John